A, Gọi X,y lần lượt là số mol của Mg và Al
Pthh:
Mg + H2SO4---> MgSO4 + H2
X. X. X. X
2Al + 3H2SO4---> Al2(SO4)3+3H2
Y. 1.5y. Y. 1.5y
Ta có pt:
24x + 27y= 1.95
X+1.5y=2.24/22.4=0.1
=> X=0.025, Y=0.05
%Mg= 0.025×24×100)/1.95=30.8%
%Al= 100%-30.8%=69.2%
mH2SO4= 0.025+1.5×0.05=0.1g
mH2= (0.025+0.05)×2=0.15g
C, Mdd H2SO4 = 0.1/6.5×100=1.54g
MddY= 1.54+1.95-0.15=3.34g
%MgSO4 vs %Al2(SO4)3 b tự tính nha

a) Sau phản ứng : $m_{chất\ rắn} = 18,88(gam)$

b) Bảo toàn khối lượng : 

$m_{O_2} = 20 - 18,8 = 1,12(gam)$

$n_{O_2} = 1,12 : 32 = 0,035(mol)$
$V_{O_2} = 0,035.22,4 = 0,784(lít)$

a)

PTHH: CuO + CO --t^o--> Cu + CO₂

            Fe₂O₃ + 3CO --t^o--> 2Fe + CO₂

\(n_{O\left(mất.đi\right)}=\dfrac{50-48,4}{16}=0,1\left(mol\right)\)

n_CO = n_{O(mất đi)} = 0,1 (mol)

=> V_CO = 0,1.22,4 = 2,24 (l)

b)

n_CO2 = n_CO = 0,1 (mol)

PTHH: Ca(OH)₂ + CO₂ --> CaCO₃ + H₂O

                               0,1---->0,1

=> \(m_{CaCO_3}=0,1.100=10\left(g\right)\)

a) \(2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\)

b)

\(n_{KClO_3} = \dfrac{36,75}{122,5} = 0,3(mol)\)

Theo PTHH : 

\(n_{KCl} = n_{KClO_3} = 0,3(mol)\\ \Rightarrow m_{KCl} = 0,3.74,5 = 22,35(gam)\\ \Rightarrow m_{O_2} = m_{KClO_3} - m_{KCl} = 14,4(gam)\)

c)

Bảo toàn khối lượng : 

\(m_{O_2} = 25 - 15,4 = 9,6(gam)\\ \Rightarrow n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)\\ n_{KClO_3} = \dfrac{2}{3}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{KClO_3} = 0,2.122,5 = 24,5(gam)\\ \%m_{tạp\ chất}= \dfrac{25-24,5}{25}.100\% = 2\%\)

\(a.\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(b.\)

\(n_{KClO_3}=\dfrac{36.75}{122.5}=0.3\left(mol\right)\)

\(\Rightarrow n_{O_2}=\dfrac{3}{2}n_{KClO_3}=\dfrac{3}{2}\cdot0.3=0.45\left(mol\right)\)

\(m_{O_2}=0.45\cdot32=14.4\left(g\right)\)

\(m_{KCl}=0.3\cdot74.5=22.35\left(g\right)\)

\(c.\)

\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\)

\(a.............a\)

\(m_{Cr}=m_{KCl}+m_{tc}=25-122.5a+74.5a=15.4\left(g\right)\)

\(\Rightarrow a=0.2\)

\(m_{O_2}=\dfrac{3}{2}\cdot0.2\cdot32=9.6\left(g\right)\)

\(m_{KClO_3}=0.2\cdot122.5=24.5\left(g\right)\)

\(m_{tc}=25-24.5=0.5\left(g\right)\)

\(\%m_{Tc}=\dfrac{0.5}{25}\cdot100\%-2\%\)

2KClO3 ---> 2KCl +3O2

nKClo3 = 24,5/122,5 = 0,2 mol

nKCl = nKClo3 =0,2 mol

m Kcl = 0,2 x 74,5 = 14,9g

no2 = 0,2x3:2 = 0,3mol

Vo2 = n.22,4 = 6,72 lít

\(V_{O_2}=n_{O_2}.22,4=0,33.22,4=7,392\left(l\right)\)

\(m_{CaCO_3}=400\cdot90\%=360\left(g\right)\)

\(m_{trơ}=400-360=40\left(g\right)\)

\(n_{CaCO_3}=\dfrac{360}{100}=3.6\left(mol\right)\)

\(a.\)

\(n_{CaCO_3\left(pư\right)}=3.6\cdot75\%=2.7\left(mol\right)\)

\(CaCO_3\underrightarrow{^{^{t^0}}}CaO+CO_2\)

\(2.7........2.7...........2.7\)

\(m_X=m_{CaO}+m_{CaCO_3\left(dư\right)}+m_{trơ}=2.7\cdot56+\left(3.6-2.7\right)\cdot100+40=281.2\left(g\right)\)

\(b.\)

\(\%CaO=\dfrac{2.7\cdot56}{281.2}\cdot100\%=53.77\%\)

\(V_{CO_2}=2.7\cdot22.4=60.48\left(l\right)\)

\(m_{CaCO_3}=90\%.400=360\left(g\right)\\ \rightarrow n_{CaCO_3}=\dfrac{360}{100}=3,6\left(mol\right)\)

PTHH: CaCO₃ --t^o--> CaO + CO₂

             3,6 ----------> 3,6 -----> 3,6

\(\rightarrow n_{CaO}=3,6.75\%=2,7\left(mol\right)\\ \rightarrow n_{CaCO_3\left(chưa.pư\right)}=3,6-2,7=0,9\left(mol\right)\)

\(\rightarrow m_X=0,9.100+2,7.56=241,2\left(g\right)\\ \%m_{CaO}=\dfrac{0,9.100}{241,2}=37,31\%\)

\(V_Y=V_{CO_2}=3,6.75\%.22,4=60,48\left(l\right)\)

\(m_{CaCO_3}=\dfrac{400\cdot90\%}{100\%}=360g\Rightarrow n_{CaCO_3}=\dfrac{360}{100}=3,6mol\)

\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)

3,6           3,6       3,6

Thực tế: \(n_{CaO}=3,6\cdot75\%=2,7mol\)

\(\Rightarrow m_{CaO}=2,7\cdot56=151,2g\)

a) \(n_{CaCO_3\left(bđ\right)}=\dfrac{400.90\%}{100}=3,6\left(mol\right)\)

=> \(n_{CaCO_3\left(pư\right)}=\dfrac{3,6.75}{100}=2,7\left(mol\right)\)

PTHH: CaCO₃ --t^o--> CaO + CO₂

              2,7---------->2,7---->2,7

=> m_X = m_{đá vôi} - m_CO2 = 400 - 2,7.44 = 281,2 (g)

b) 

m_CaO = 2,7.56 = 151,2 (g)

=> \(\%CaO=\dfrac{151,2}{281,2}.100\%=53,77\%\)

V_CO2 = 2,7.22,4 = 60,48 (l)

\(n_{KClO_3}=\dfrac{12,25}{122,5}=0,1\left(mol\right)\)

PTHH: 2KClO₃ --to--> 2KCl + 3O₂ - pư phân huỷ

             0,1                    0,1      0,15

\(\rightarrow m_{KCl}=0,1.74,5=7,45\left(g\right)\)

Zn + 2HCl ---> ZnCl2 + H2 

0.2-→0.4------→0.2--→0.2 (mol)

nZn = 11,2\56 = 0.2(mol)

mZnCl2 = n*M = 0.2*127 = 25.4(g)

VH2(đktc) = n*22.4 = 0.2*22.4 = 4.48(l)

mHCl = n*M = 0.4*36.5 = 14.6(g)

C% =14,6\146*100% = 10(%)

\(a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{11,2}{65}=0,17\left(mol\right)\\ TheoPT:n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,17\left(mol\right)\\ m_{ZnCl_2}=0,17.136=23,12\left(g\right)\\ V_{H_2}=0,17.22,4=3,808\left(l\right)\\ c.n_{HCl}=2n_{Zn}=0,34\left(mol\right)\\ C\%_{HCl}=\dfrac{0,34.36,5}{146}.100=8,5\%\)